Topic 6 Flashcards

1
Q

Mutations in which organ has the greatest evolutionary effect?

A

the gonads

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2
Q

What are the two types of mutations?

A

Germinal: only mutations in the germ cells will be transmitted to the progeny

Somatic mutations: will impact the individual where the mutation occurs but are evolutionarily irrelevant

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3
Q

What is the reigning paradigm of mutations?

A

Mutations can occur in any cell at any time, and their occurrence is ‘random’.

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4
Q

True or false? Mutagenic processes are at a constant tug-of-war with eachother.

A

True

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5
Q

DNA is a remarkably stable molecule, but it can still be damaged due to chemical or physical stress on the double helix.
What are the five types of damage?

A

(1) Abasic sites (loss of nucleotide, not backbone)
(2) Base mismatches
(3) Modified bases
(4) Inter- and intra-strand crosslinks
(5) Double stranded breaks (DSBs)

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6
Q

What is the range of genetic mutations?

A

Causes everything from point mutation (1bp affected) to structural changes affecting the whole chromosomes (deletions, inversions, duplications, translocations)

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7
Q

Compare spontaneous and induced mutations.

A

spontaneous mutations are mostly from replication errors (polymerase-induced mistakes) followed by defects in the DNA repair mechanism. The rate of spontaneous mutations is very low and varies according to gene size, domains on the chromosome, genome characteristics, and cell age.

Induced mutations (known as chemical and physical agents or mutagens) have a higher frequency and include base analogs, hydroxylating agents, alkylating agents, deaminating agents, intercalating agents, UV radiation, and ionizing radiation.

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8
Q

What are point mutations and where do they take place?

A

Single base pair changes or single nucleotide polymorphism (SNP) that can occur anywhere in the genome (coding regions, intergenic regions, non-coding regions of genes like promoters, UTRs, or introns)

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9
Q

What are the three types of point mutations?

A

(1) Base substitutions
(2) Base deletion
(3) Base insertion

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10
Q

What are the kinds of base substitution point mutations?

A

Transition:
purine replaced by a purine (A-G, G-A)
pyrimidine replaced by a pyrimidine (C-T, T-C)

Transversion:
purine replaced by a pyrimidine (A-C, A-T, G-C, G-T)
Pyrimidine replaced by a purine (C-A, C-G, T-A, T-G)

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11
Q

Name synonymous and non-synonymous mutations.

A

Synonymous:
Silent mutations- encodes the same amino acid, usually in the third degenerate position, with no effect on the phenotype

Non-synonymous:
Missense Mutations- codes for a different amino acid
Nonsense mutations- changes to a STOP codon, likely severe

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12
Q

What will cause for a missense mutation to either retain protein function or not?

A

Chemically similar– conservative, may retain protein function
Chemically different– non-conservative, more likely to affect protein function

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13
Q

Describe a frameshift mutation.

A

Removal or addition of base pairs disrupts the triplet reading frame.
The result is the translation of an abnormal series of aa downstream from the indel and the increased chances of a premature STOP codon producing a truncated protein.
Indels (insertions/deletions) that are multiple of three reconstitute the frame downstream of the last mutation site

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14
Q

Explain functional consequences of mutations in the non-coding regions of DNA.

A

Point mutations in regulatory regions required for gene expression (promoters, enhancers, etc.) or that are involved in mRNA processing and stability (splice sites, UTRs) may ultimately affect how much protein is made

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15
Q

A mutation in the Shine-Delgarno sequence will effect
(a) Prokaryotic transcription
(b) Eukaryotic transcription
(c) Prokaryotic translation
(d) Eukaryotic translation
(e) none of the above

A

Prokaryotic translation

This ribosomal binding site in bacterial messenger RNA became known as the Shine-Dalgarno (SD) sequence enables initiation of protein synthesis by aligning the ribosome with the start codon.

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16
Q

What mutation is caused by error in replication?

A

DNA slippage causes indels during replication– usually of repeated regions

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17
Q

What does slippage mutations result in?

A

The daughter strand gets looped and that part of the template is repeated

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18
Q

What is the result of CGG expansion in Fragile X syndrome?

A

The methylation of 5’ regulatory sequences and transcription silencing

Note: slippage is behind an important class of mutations defined by trinucleotide repeats that explains over 40 inherited neurological diseases such as Fragile X syndrome (CGG repeats in the FMR1 gene) and Huntington Disease (CAG repeats in the huntington gene)

19
Q

How does energetically favourable hydrogen bonding ensure DNA strand complementarity? How do nucleotide mispairing mutations occur?

A

The geometric correspondence between purine and pyrimidines allows the specific formation of weak hydrogen bonds. Unfavourable associations are prevented because hydrogen atoms are either too far away or repulsed by close proximity.

Alteration of the chemical geometry (isomerization) of nucleotides can change the favourable hydrogens suitable for pairing, and break complementarity rules.

20
Q

Mismatches in the replicating DNA strand is correct by…

A

DNA polymerase proofreading (5’ to 3’ exonuclease activity and by DNA repair mechanisms)

21
Q

Describe the three cellular environmental mutation causes: depurination, deamination, and oxidative damage

A

Depurination: loss of purine base (G or A), efficiently repaired by endogenous DNA repair mechanisms

Deamination: removal of an amine (-NH2) group from C, A and G, alters base pairing (with the exception of Xanthine) and leads to mutations during replication

Oxidative damage: reactive Oxygen species (ROS) are byproducts of aerobic metabolism in mitochondria induced by ionizing radiation. ROS agents can damage DNA in several different ways, leading to replication stall or mismatch pairings

22
Q

Define mutagen

A

a chemical or physical force that can increase the mutation rare above mu (p)

23
Q

What are three functions of mutagens?

A

Mutagens can:
(1) replace a base pair in the DNA strand
(2) chemically alter a base pair leading to a mismatch
(3) damage a base pair such that it cannot base pair with any other nucleotide

24
Q

Give some examples of chemical and physical types of mutagen agents.

A

Chemical agents:
alkylating agents, ROS, intercalating agents, DNA adducts, base analogs

Physical agents:
UV/ionizing radiation

25
Q

Which of the following is not an example of DNA damage likely to cause a mutation?
(a) abasic site
(b) base analogs
(c) intercalation
(d) backbone break
(e) none of the above

A

None of the above: mutations are all or none

26
Q

How do physical mutagens induce mutations?

A

Covalently links neighbouring pyrimidines and forms pyrimidine dimers
Can’t be recognized by DNA polymerase– stops DNA replication
Can be repaired but often results on transition mutations

27
Q

How do you determine if a chemical is a mutagen or carcinogen?

A

The frequency of mutant animals that can survive after exposure to a mutagen reflects the number of times the reversion mutation occurred in the same position in the same gene because of exposure to a mutagen. This frequency is a measure of mutagenic potential of mutagens.

28
Q

Summarize the Ames test in salmonella.

A

Salmonella his- is a point mutation in a gene necessary for histidine synthesis– an essential amino acid.
A second mutation in the same site may revert the base pair back to the wild type sequence such that the strain can again synthesize histidine, this reverts the phenotype back to wildtype.
If this reversion occurs, Salmonella cells will be able to grow on plates without supplemented histidine.
By treating cells with different concentrations of potential carcinogens and determining the frequency of revertants, the Ames test can quantify the relative mutagenic strength of different compounds.

29
Q

How do you determine Mutagenicity Ratio (MR)?

A

MR = total number of revertants / number of spontaneous revertants

The control plate shows spontaneous mutation rate
The test plate displays the total (spontaneous + induced mutation rate)

30
Q

What are 6 major DNA repair systems?

A

(1) Base Excision Repair (BER)
(2) Nucleotide Excision Repair (NER)
(3) Mismatch repair (MMR)
(4) Translesion Synthesis (TLS)
(5) Homologous Recombination Repair (HR)
(6) Non-homologous End-Joining (NHEJ)

31
Q

What are the four components in the DNA repair system?

A

(1) Surveillance: error detection post replication
(2) Excision: enzyme removes or alters the bp(s) involved
(3) Polymerization: uses undamaged template/homolog to re-polymerase the removed bases
(4) Strand ligation: reconnects any sugar-sugar bonds in the repaired strands

32
Q

Describe direct repair mechanisms.

A

Some DNA damage can be directly reverted using specific enzymes that identify altered specific nucleotides

33
Q

Describe base excision repair mechanims.

A

Removes and replaces damaged bases due to alkylation, oxidation and deamination
Relies on complementarity on the non-affected strand to correct the mistake

(1) DNA glycosylases remove the base (creates an abasic site)
(2) the damaged strand is nicked by an endonuclease to allow for the removal of the sugar by a phosphodiesterase or DNA polymerase
(3) DNA polymerase adds the correct complementary base (directly or as a longer repair flap)
(4) A ligase reconnects the DNA strand

34
Q

Describe nucleotide excision repair mechanisms for large damage affecting multiple base pairs, bulky adducts, and pyrimidine dimers.

A

(1) Damage detection
(2) strand separation
(3) incision
(4) excision
(5) polymerization
(6) Ligation

35
Q
A

Autosomal recessive

Parents are AA and aa
because mutant is recessive to one parent is guaranteed to be homozygous recessive and the mutation is rare so we assume the other parent is homozygous dominant.

100% probability that V-1 is heterozygous

36
Q

Describe mismatch repair.

A

(1) Detection
Mismatches during replication cause distortions in the DNA strand (loops) that are recognized; must detect new strand vs template strand
(2) Incision- nick to allow excision
(3) Excision- helicase unwinds and region around mismatch removed by exonucleases
(4) synthesis- DNA polymerase fills the gap
(5) Ligation- ligase closes the incisions

37
Q

DNA repair mechanisms overview

A

Translation synthesis was not covered
Four main types: base excision repair, nucleotide excision repair, mismatch repair, and nonhomologous end joining.
Simplest is basic excision repair: one of the bases has a small modification to it, fairly simple repair because the repair mechanism can know the correct base
More serious/awkward damage: nucleotide excision repair, takes out the whole nucleotide including the backbone, or a bulky adduct.
Likely happening with DNA polymerase: mismatch repair, proteins and functions in the polymerase itself
Most serious: double stranded breaks, two ways of repair, the best is HR but also uses NHEJ, broken DNA ends erode quickly so this is an emergency repair

38
Q

AutoYou perform an Ames test and see 5 revertants in the control and 20 revertants on the test plate. What is the mutagenicity ratio?

A

MR = Total/ control
20/5=4

39
Q

Why is a double strand break (DSB) so serious?

A

Cannot use template to repair
Repair is critical: can lead to chromosomal abnormalities or cell death.

Mutations in DSB repair genes lead to a series of hereditary neurodegenerative, developmental disorders, immunnodeficiencies and cancer predispositions

40
Q

Compare homologous recombination and non-homologous end joining DBS repair

A

Do not need to know the complex steps and all the details, just know how HR is different from NHEJ

41
Q

Describe Non-homologous end joining DBS repair.

A

(1) Detection (surveillance)
proteins bind each end of the break to suppress further damage

(2) Strand Resection
Recruitment of kinase and nuclease proteins
remove 5’ and 3’ overhangs: makes blunt ends

(3) Polymerization
DNA polymerase fills ends for ligation

(4) Ligation
Ligase ligates the two ends

42
Q

Describe homologous recombination DBS repair.

A

Note: requires an undamaged template from a sister chromatid or homologous chromosome

(1) Detection

(2) Strand Resection
Creates 3’ overhands (sticky ends) required for strand exchange

(3) Strand exchange/invasion
3’ overhang invades template strands
recombinase mediated
Forms a D-loop (displacement loop)

(4) Polymerization
Using the homolog/sister as template, the invading strand is extended by polymerases

These intermediates can be either processed by:
Synthesis dependent strand annealing pathway (SDSA)
Double strand break repair pathway (BSBR)

43
Q

The intermediates of homologous recombination DBS repair can then be processed by either
(1) Synthesis Dependent Strand Annealing Pathway (SDSA)
(2) Double Strand Break Repair Pathway (DSBR)

Describe the first.

A

DNA helicases break invading off the homologous template after polymerization and original strands re-anneal.
Gaps are filled by DNA polymerase and strands reconnected by DNA ligase
There is no chance of strand exchange in this scenario.

44
Q

The intermediates of homologous recombination DBS repair can then be processed by either
(1) Synthesis Dependent Strand Annealing Pathway (SDSA)
(2) Double Strand Break Repair Pathway (DSBR)

Describe the second.

A

(1) Second strand invades, associates with template
This forms a two crossover intermediate structure referred as a double Holliday Junctions

(2) Resolution of intermediates
Endonucleases cleave HJs to release recombined or non-recombined products

(3) Polymerization and ligation
Polymerases fill gaps between the strands and DNA ligases connect the cleaved strands